# What are common mistakes students make when finding the limit of an infinite sequence?

Oct 1, 2017

$\textcolor{red}{\frac{\infty}{\infty} = 1}$ and $\textcolor{b l u e}{\infty - \infty = 0}$ are typical mistakes. When you think of limitation, be sure to concider the circumstances.

#### Explanation:

When you think about the limitation, ${\lim}_{n \to \infty} {a}_{n}$, don't forget that
$\infty$ is NOT a infinitely large $\textcolor{b l u e}{\text{number}}$, BUT a $\textcolor{red}{\text{situation}}$ $n$ is increasing infinitely.

[$\textcolor{red}{\text{example1}}$] Where does ${n}^{2} / n$ go when $n$ goes to infinity?

Here we must evaluate ${\lim}_{n \to \infty} {n}^{2} / n$
It is true ${\lim}_{n \to \infty} n = \infty$ and ${\lim}_{n \to \infty} {n}^{2} = \infty$, but if you think
${\lim}_{n \to \infty} {n}^{2} / n = \frac{\infty}{\infty} = 1$ you are wrong.

The situation shows ${n}^{2}$ is growing much faster than $n$ and thus the limitation cannot be $1$.

To solve this, you first need to $\textcolor{red}{\text{reduce the fraction}}$ and then let $n \to \infty$.
${\lim}_{n \to \infty} {n}^{2} / n = \textcolor{red}{{\lim}_{n \to \infty} \frac{n}{1} = {\lim}_{n \to \infty} n = \infty}$
is the correct way.

[$\textcolor{b l u e}{\text{example2}}$] What is ${\lim}_{n \to \infty} \left(n - \sqrt{{n}^{2} + 2 n}\right)$?

You might think:
${\lim}_{n \to \infty} n = \infty$ and ${\lim}_{n \to \infty} \sqrt{{n}^{2} + 2 n} = \infty$ and $\infty - \infty = 0$・・・Just wait! This is wrong again.

Here we should $\textcolor{b l u e}{\text{ratonalize}}$ $n - \sqrt{{n}^{2} + 2 n}$.
${\lim}_{n \to \infty} \left(n - \sqrt{{n}^{2} + 2 n}\right) = {\lim}_{n \to \infty} \frac{\left(n - \sqrt{{n}^{2} + 2 n}\right) \left(n + \sqrt{{n}^{2} + 2 n}\right)}{n + \sqrt{{n}^{2} + 2 n}}$
$= \textcolor{red}{{\lim}_{n \to \infty} \frac{- 2 n}{n + \sqrt{{n}^{2} + 2 n}}}$ ・・(A)

(A) is $\frac{\infty}{\infty}$ again, so divide both numerator and denominator by $n$.

(A)$= \textcolor{b l u e}{- {\lim}_{n \to \infty} \frac{2}{1 + \sqrt{1 + \frac{2}{n}}}}$
$= - \frac{2}{1 + 1} = - 1$ and this is the goal.

In general, $\frac{\infty}{\infty}$, $\frac{0}{0}$, $\infty \times 0$, $\infty - \infty$, ${1}^{\infty}$ or ${\infty}^{0}$ typed limitations are called $\textcolor{g r e e n}{\text{indeterminate form}}$ and they cannot be calculated like numbers. They should be treated carefully.